MATHS REVISION TEST NO. 2 ( FOR CBSE STD IX )

REVISION TEST 2 1] The radius of a circle is 13 cm and the length of one of its chords is 10 cm. Find the distance of the chord from the centre. 2] In the above figure O is the centre of the circle and L BOD = 1600. Find angles C and A marked with letters x and y respectively. (please mark the centre as O ) 3]The angles of a quadrilateral are (x – 20 )0, (x + 80)0, (2x – 80)0 and (2x + 20 )0 . Find all angles of the quadrilateral and show that this quadrilateral is a parallelogram. 4] In the following figure , ABC represents a triangular plot where AB = 60 m and BC = 40 m. CP the perpendicular distance from C to AB is 32 m. Find AQ, the perpendicular distance from A to BC. 5] The sides of a triangle are 6 cm , 10 cm and 14 cm. Find the area of the triangle 6] In the given circle , L BAC = 600 and L BCA = 200. Find L ADC 7] Using ruler and compass only, construct triangle ABC in which L B = 600 , L C = 450 and with perimeter 11 cm 8] In the given figure, P , Q and R are the mid-points of AB , BC and CA respectively. Given PQ = 3cm , QR = 4 cm and PR = 5cm. (i) Find the perimeter of triangle ABC (ii) show that ABC is a right angled triangle. 9] Find four solutions for the equation, 10] Draw the axes and plot the points A (-1 , -1) , B ( 4 , -1) and C (5 , 3). D is another point such that ABCD is a parallelogram. Find the coordinates of D 11] using a ruler and compass only, construct triangle ABC in which AB = 6 cm, L A = 300 and L B = 1350 ANSWERS 1] Join the centre of the circle to the mid-point of the chord and get a right triangle Applying Pythagoras theorem in this right triangle, distance d = 2] L x = 160 / 2 = 800 ( angle subtended by an arc at the centre = twice the angle subtended by the same arc at the remaining part of the circumference) Now L y = 180 – 80 = 1000 ( sum of angle x and y is 180 because they are the opposite angles of a cyclic quadrilateral ) 3] The sum of the angles of any quadrilateral is 360. So (x – 20) + (x + 80) + (2x – 80) + (2x + 20) = 360 Solving the above equation , we get x = 60. Substituting for x , the angles will be 40 , 140 , 40 and 140 degrees. Hence this will be parallelogram ( since the opposite angles are equal) 4] Area of the triangle = ½ x 60 x 32. Area of the same triangle = ½ x 40 x AQ So ½ x 60 x 32 = ½ x 40 x AQ Solving the above or cancelling both sides , we AQ = 48 cm 5] semi-perimeter= s = (6 + 10 +14)/2 = 15 cm So Area A = = = 15 sq. cm 6] L B = 180 – (20 + 60 ) = 100 degrees. So L D = 180 – 100 = 80 degrees. (Since ABCD is a cyclic quadrilateral and its opposite angles are supplementary) 7] Draw a base line LM of length 11 cm, which should be the perimeter of the triangle ABC. Constrict 60 degrees at L and 45 degrees at M and make them meet at a point A, the supposed vertex of triangle ABC . Join LA and construct its perpendicular bisector so as to meet the base line at B. B will be the second vertex of the required triangle. Join MA and construct its perpendicular bisector cutting the base line at B, the third vertex of the required triangle. Join ABC. That will be the vertices of the required triangle. 8] (a) In the figure BC = 2 PR ; CA = 2 PQ and AB = 2 QR ( Mid point theorem) Therefore , BC =2 x 5 = 10 cm , AC = 2 x 3 = 6 cm and AB = 2 x 4 = 8 cm. (b) Also AC2 + AB2= 62 + 82 = 36 + 64 = 100 = BC2 Therefore ABC will be a right triangle with hypotenuse BC or right angle at A. 9] Putting x = 0 , y =3 . Hence ( 0 , 3 ) will be a solution. When x = 2 , ½ x 2 + y = 3; 1 + y = 3 or y =2. So (2, 2) will be another solution of the equation . Putting any other values for x =(eg, 4 , 6 , 8 etc ), you can find the other solutions. 10] Draw the axes and mark the three points A , B , and C first. Check how the point C is displaced from the point B. It will be found that C has been shifted 1 unit to the right and 4 units up. Using the same way obtain point D counting 1 unit to the right and 4 units up. Point D will be (0, 3) 11] Draw the base, AB first taking 6 cm. Construct 60 degrees at A and bisect it to get a 30 degree line. . construct 45 degrees at the right side of B so that the angle at the left side will be ( 180 – 45 ) = 135 degrees. Find the point of intersection of the lines making 30 deg and 135 deg and mark it as C. Complete triangle ABC